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          状态反馈思考与仿真2——平衡点与随动系统
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        <h1 id="平衡点fixed-point">平衡点（Fixed point）</h1>
<p>在<a target="_blank" rel="noopener" href="https://xzhmars.gitee.io/2021/09/10/状态反馈思考与仿真/">上一篇博文</a>中，随着状态反馈的加入，原本不稳定的系统变得稳定，在初始状态不为0的条件下，系统状态均可收敛于0；此外，当阶跃随着<span class="math inline">\(v\)</span>加入后，系统也可稳定，但却没有收敛0点，这是为什么？</p>
<p>先引入平衡点的概念，对于一个<span class="math inline">\(n\)</span>​阶系统，其状态变量为<span class="math inline">\(x_1,x_2...x_n\)</span>​。当<span class="math inline">\(\dot{x}_1=0,\dot{x}_2=0...\dot{x}_n=0\)</span>​时，<span class="math inline">\(x_1=x_{1f},x_2=x_{2f}...x_n=x_{nf}\)</span>​，<span class="math inline">\(x_{1f},x_{2f}...x_{nf}\)</span>​就被称为系统的平衡点。</p>
<p>一个系统，在静止条件下，一般都会收敛于平衡点(why?)，我们设计的状态反馈，会促使系统在动态条件下收敛于平衡点(why?)。例如，还是我们之前的小车模型：</p>
<p><img src="image-20210911113226022.png" alt="image-20210911113226022" style="zoom:15%;" /></p>
<p>设计<span class="math inline">\(\lambda_{1,2}=-1，-2\)</span>，可得<span class="math inline">\(k_1 = 2,k_2=3\)</span>，加入状态反馈后，系统微分方程： <span class="math display">\[
\left\{\begin{array}{l}
\dot{x}_{1}=x_{2} \\
\dot{x}_2 = -2x_1 - 3x_2+u
\end{array}\right.
\]</span> 令<span class="math inline">\(\dot{x}_1=0,\dot{x}_2=0\)</span>，可得<span class="math inline">\(x_{1f} = 0,x_{2f}=0\)</span>​；所以系统会收敛于0点​。对于两个小车模型：</p>
<p><img src="image-20210911163815643.png" alt="image-20210911163815643" style="zoom:15%;" /></p>
<p>加入状态反馈，设计<span class="math inline">\(\lambda_{1,2}=-2 \pm 2 \sqrt{3}i,\lambda_{3,4}=-10\)</span>，系统微分方程： <span class="math display">\[
\left\{\begin{array}{l}
\dot{x}_{1}=x_{2} \\
\dot{x}_2 = -96x_1 - 24x_2+ 80x_3+16.8x_4\\
\dot{x}_3 = x_{4}\\
\dot{x}_4 = 100x_1 - 100x_3
\end{array}\right.
\]</span> 同理令<span class="math inline">\(\dot{x}_1=0,\dot{x}_2=0,\dot{x}_3=0,\dot{x}_4=0\)</span>，有<span class="math inline">\(x_{1f} = 0,x_{2f}=0，x_{3f} = 0,x_{4f}=0\)</span>.</p>
<h1 id="轨迹跟踪">轨迹跟踪</h1>
<p>在配置状态反馈后，系统收敛于状态变量等于0的点，但如果想要构成一个随动系统应该怎么设计呢？</p>
<p>可以这样思考，既然收敛于状态变量等于0的点，不如替换状态变量为误差，让误差趋近于0，这样就可以完成随动系统的设计。</p>
<p>对于单个小车模型，我们期望小车停在<span class="math inline">\(x_1 = x_{1d}\)</span>的位置，所以引入误差：<span class="math inline">\(e=x_{1d}-x_1\)</span>​，求导，我们获得第一个微分方程： <span class="math display">\[
\dot{e} = -\dot{x}_1 = -x_2
\]</span> 由<span class="math inline">\(u=m\ddot{x}_1=\dot{x}_2\)</span>得： <span class="math display">\[
\dot{x}_2 = u
\]</span> 得到新的状态方程： <span class="math display">\[
\begin{bmatrix}
\dot{e}  \\
\dot{x}_2 
\end{bmatrix}
=
\begin{bmatrix}
0 &amp; -1 \\
0 &amp; 0 
\end{bmatrix}
x +
\begin{bmatrix}
0 \\
1 
\end{bmatrix}u
\]</span> 对于这个系统我们依然考察其平衡点：令<span class="math inline">\(\dot{e}=\dot{x}_2=0\)</span>有： <span class="math display">\[
\left\{\begin{array}{l}
e_f = ?\\
x_{2f} = 0
\end{array}\right.
\]</span> 无法解出<span class="math inline">\(e_f\)</span>的具体值，代表<span class="math inline">\(e\)</span>没有稳定的点。因此此时我们加入状态反馈，把系统的极点配置到<span class="math inline">\(\lambda_{1,2}=-1，-2\)</span>。可得<span class="math inline">\(k_{1,2} = -2,3\)</span>​.新系统变为 <span class="math display">\[
\begin{bmatrix}
\dot{e}  \\
\dot{x}_2 
\end{bmatrix}
=
\begin{bmatrix}
0 &amp; -1 \\
2 &amp; -3 
\end{bmatrix}
x +
\begin{bmatrix}
0 \\
1 
\end{bmatrix}u
\]</span> 再次考察加入状态反馈的平衡点：令<span class="math inline">\(\dot{e}=\dot{x}_2=0\)</span>有： <span class="math display">\[
\dot{x}_2 = 2e -3x_2+u
\]</span></p>

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